/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 If a certain amount of ideal gas... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a certain amount of ideal gas occupies a volume \(V\) at STP on earth, what would be its volume (in terms of \(V\) ) on Venus, where the temperature is \(1003^{\circ} \mathrm{C}\) and the pressure is 92 atm?

Short Answer

Expert verified
The volume on Venus is approximately 0.0508V.

Step by step solution

01

Understanding STP Conditions

At STP (Standard Temperature and Pressure) on Earth, the conditions are 0°C (273 K) and 1 atm of pressure. We use these standard values to calculate changes when conditions change.
02

Identify the Given Conditions on Venus

The temperature on Venus is given as 1003°C, which needs to be converted to Kelvin for calculations. Thus, the temperature is 1003°C + 273 = 1276 K. The pressure on Venus is given as 92 atm.
03

Apply the Ideal Gas Law

The ideal gas law in terms of volume is given by \( PV = nRT \). For a constant amount of gas \(n\), the equation can be converted into the relationship \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \), where subscript 1 indicates the initial conditions (STP) and subscript 2 indicates Venus conditions.
04

Set Up the Equation for Volume Change

Using the relationship \( \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \), substitute the known values: \( P_1 = 1 \) atm, \( V_1 = V \), \( T_1 = 273 \) K, \( P_2 = 92 \) atm, and \( T_2 = 1276 \) K. The equation becomes: \( \frac{1 \cdot V}{273} = \frac{92 \cdot V_2}{1276} \).
05

Solve for the New Volume on Venus

Rearrange the equation to solve for \( V_2 \): \( V_2 = \left(\frac{1 \cdot V}{273}\right) \times \frac{1276}{92} \). Cancel and simplify: \( V_2 = V \left( \frac{1276}{92 \times 273} \right) \).
06

Calculate and Simplify the Expression

Perform the calculation: \( V_2 = V \times \frac{1276}{25116} \approx V \times 0.0508. \) This states that the gas will occupy approximately 5.08% of its original volume \( V \) at STP.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

STP conditions
The concept of Standard Temperature and Pressure (STP) is crucial for comparing gas behavior across different conditions. At STP, temperature is standardized at 0°C, which is equivalent to 273 Kelvin (K), and the pressure is set at 1 atmosphere (atm). These conditions ensure a common ground when studying gases, as they provide a baseline to calculate how a gas might behave when subjected to different temperatures and pressures.

The Ideal Gas Law, a key principle in chemistry, utilizes these standard conditions through the formula: \[ PV = nRT \] where \( P \) is pressure, \( V \) is volume, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, and \( T \) is temperature in Kelvin. When comparing changes in conditions, the ideal gas law can be adapted to: \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] helping us understand how a gas's volume changes from one environment to another, such as from Earth to Venus.
Volume calculation
Volume calculation under varying conditions using the Ideal Gas Law involves understanding how changes in pressure and temperature affect how much space a gas will occupy. Given the formula \[ \frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \] we see how initial conditions (subscript 1) and changed conditions (subscript 2) relate.

For example:
  • If initial conditions are at STP (with \( P_1 = 1 \) atm, \( V_1 = V \), and \( T_1 = 273 \) K), we can identify the new environment's conditions.
  • On Venus, with a much higher temperature of 1276 K and pressure of 92 atm, the equation enables the calculation of the new volume \( V_2 \).
Setting up the equation: \[ \frac{1 \cdot V}{273} = \frac{92 \cdot V_2}{1276} \] allows us to solve for \( V_2 \): \[ V_2 = V \times \frac{1276}{92 \times 273} \] This leads to discovering that the gas occupies only about 5.08% of its initial STP volume when on Venus.
Pressure and temperature on Venus
Venus presents conditions that are drastically different from those on Earth, making it an interesting subject for studying gas behavior. The planet's surface is extremely hot, with temperatures reaching \( 1003^{\circ} \text{C} \) (converting to 1276 K for calculations).

The surface pressure is extremely high at 92 atm, which significantly compresses gases compared to the 1 atm at STP on Earth.
  • This high pressure tends to decrease the volume a gas can occupy.
  • Conversely, the high temperature usually increases the energy and expands the volume of a gas.
The balance between these two opposing factors (high pressure vs. high temperature) is what defines the final volume of a gas transported to Venus from Earth. In effect, on Venus, the immense pressure more than offsets the expansion due to heat, resulting in a significant reduction of the gas volume.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A large cylindrical tank contains \(0.750 \mathrm{~m}^{3}\) of nitrogen gas at \(27^{\circ} \mathrm{C}\) and \(1.50 \times 10^{5} \mathrm{~Pa}\) (absolute pressure). The tank has a tight-fitting piston that allows the volume to be changed. What will be the pressure if the volume is decreased to \(0.480 \mathrm{~m}^{3}\) and the temperature is increased to \(157^{\circ} \mathrm{C} ?\)

An automobile tire has a volume of \(0.0150 \mathrm{~m}^{3}\) on a cold day when the temperature of the air in the tire is \(5.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.02 atm. Under these conditions, the gauge pressure is measured to be 1.70 atm (about \(25 \mathrm{lb} /\) in. \(^{2}\) ). After the car is driven on the highway for \(30 \mathrm{~min}\), the temperature of the air in the tires has risen to \(45.0^{\circ} \mathrm{C}\) and the volume to \(0.0159 \mathrm{~m}^{3}\). What is the gauge pressure at that time?

One type of gas mixture used in anesthesiology is a \(50 \% / 50 \%\) mixture (by volume) of nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) and oxygen \(\left(\mathrm{O}_{2}\right),\) which can be premixed and kept in a cylinder for later use. Because these two gases don't react chemically at or below 2000 psi, at typical room temperatures they form a homogeneous single- gas phase, which can be considered an ideal gas. If the temperature drops below \(-6^{\circ} \mathrm{C},\) however, \(\mathrm{N}_{2} \mathrm{O}\) may begin to condense out of the gas phase. Then any gas removed from the cylinder will initially be nearly pure \(\mathrm{O}_{2}\). As the cylinder empties, the proportion of \(\mathrm{O}_{2}\) will decrease until the gas coming from the cylinder is nearly pure \(\mathrm{N}_{2} \mathrm{O}\). In a test of the effects of low temperatures on the gas mixture, a cylinder filled at \(20.0^{\circ} \mathrm{C}\) to 2000 psi (gauge pressure) is cooled slowly and the pressure is monitored. What is the expected pressure at \(-5.00^{\circ} \mathrm{C}\) if the gas remains a homogeneous mixture? A. 500 psi B. 1500 psi C. 1830 psi D. 1920 psi

An ideal gas expands while the pressure is kept constant. During this process, does heat flow into the gas or out of the gas? Justify your answer.

In a cylinder, \(4.00 \mathrm{~mol}\) of helium initially at \(1.00 \times 10^{6} \mathrm{~Pa}\) and \(300 \mathrm{~K}\) expands until its volume doubles. Compute the work done by the gas if the expansion is (a) isobaric and (b) adiabatic. (c) Show each process on a \(p V\) diagram. In which case is the magnitude of the work done by the gas the greatest? (d) In which case is the magnitude of the heat transfer greatest?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.